SHSAT MATH SAMPLE TEST QUESTIONS AND ANSWERS

About "SHSAT Math Sample Test Questions And Answers"

Here we are going to see some practice questions questions for SHSAT exams.

SHSAT Math Sample Test Questions and Answers - Practice questions

Question 11 :

If N = √(36 + 49), then N is

Solution :

N = √(36 + 49)

N = √85

The square root of 81 is 9.

But root 85 is greater than 81, the number between 9 and 10.

Question 12 :

Susan is 5 years older than Phen is now. In N years, Susan will be twice as old as Phen is now. If Susan is now 22 years old, what is the value of N?

Solution :

Let "x" be Phen's age

x + 5 be Susan's age

x + 5 + N = 2x ---(1)

Now Susan's age = 22 years

x + 5 = 22

x = 22 - 5 = 17

By applying the value of x in (1)

17 + 5 + N = 2(17)

22 + N = 34

N = 34 - 22

N = 12

Question 13 :

The number of integer values of n for which 1 ≤ √n ≤ 3 is

Solution :

1 ≤ √n ≤ 3

√1 = 1

√2 = 1.41....

√3 = 1.73

√4 = 2

√5 = 2....

√9 = 3

Hence 9 is number of integer values of n.

Question 14 :

In right triangle ABC, angle ACB is 90°. The number of degrees in angle BEC is

Solution :

In triangle ACB :

<ACB + <CBA + <BAC = 180

90 + <CBA + 20 = 180

<CBA = 180 - 110 = 70

Now consider the triangle CEB,

<CEB + <CBE + <BCE = 180

<CEB + 70 + 70 = 180

<CEB = 180 - 140

<CEB = 40

Question 15 :

If it is now 12:00 noon, what time was it 40 hours ago?

Solution :

40 + 12 = 52

By dividing 52 by 12, we get the quotient 4 and remainder as 4.

Now, we have to move the clock 4 hours back word from 12 : 00 noon. So we get 8 : 00 AM.

Question 16 :

The mean of all the odd integers between 6 and 24 is

Solution :

First let us list out the odd integers between 6 and 24.

7, 9, 11, 13, 15, 17, 19, 21, 23

So, there are 9 odd numbers lies between 6 and 24.

Mean = (7 + 9 + 11 + 13 + 15 + 17 + 21 + 23)/9

= 135/9

= 15

Question 17 :

Let x be an element of the set {.2, 1.2, 2.2, 3.2, 4.2}. For how many values of x is 10x/3 an integer?

Solution :

Let f(x) = 10x/3

x = 0.2

= 10(0.2)/3

= 2/3

Not integer

x = 1.2

= 10(1.2)/3

= 12/3

= 4

= integer

x = 2.2

= 10(2.2)/3

= 22/3

= Not integer

x = 3.2

= 10(3.2)/3

= 32/3

= Not integer

x = 4.2

= 10(4.2)/3

= 42/3

= integer

Hence for 2 values of x, we get integer.

Question 18 :

George has just enough money to buy 3 chocolate bars and 2 ice cream cones. For the same amount money, he could buy exactly 9 chocolate bars. For the same amount of money, how many ice cream cones could George buy?

Solution :

Let "x" and "y" be the cost of one chocolate bar and one ice cream cone.

3x + 2y = f(x) ---(1)

Cost of 9 chocolate bars = 9x = f(x) ----(2)

(1) = (2)

3x + 2y = 9x

2y = 9x - 3x

2y = 6x

y = 3x ==> x = y/3

By applying y = 3x in (1), we get

3(y/3) + 2y = f(x)

y + 2y = f(x)

f(x) = 3y

Hence, we can buy 3 ice cream cone for the same amount.

Question 19 :

ABCD and PQRS are squares, as shown. The area of PQRS is

Solution :

To find the side length pf RS, we have to use Pythagorean theorem.

In triangle DSR,

SR2 = SD2 + DR2

SR2 = 12 + 22 = 5

SR = √5

Area of the square PQRS = a2

= (√5)2 = 5

Question 20 :

If x = 10 and y = 8, what is the value
of y (3x – 2y)?

Solution :

= y(3x – 2y)

x = 10, y = 8

= 8(3(10) – 2(8))

= 8 (30 - 16)

= 8 (14)

= 112

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